Consider again the system x _ = Ax = _1 1 1 3 _ x (36) that we discussed in Example 2

Chapter 7, Problem 15

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Consider again the system x _ = Ax = _1 1 1 3 _ x (36) that we discussed in Example 2. We found there that A has a double eigenvalue r1 = r2 = 2 with a single independent eigenvector (1) = (1, 1) T , or any nonzero multiple thereof. Thus one solution of the system (36) is x(1) (t) = (1) e2t and a second independent solution has the form x(2) (t) = te2t + e2t , where and satisfy (A 2I) = 0, (A 2I) = . (37) In the text we solved the first equation for and then the second equation for . Here, we ask you to proceed in the reverse order. a. Show that satisfies (A 2I)2 = 0. b. Show that (A 2I)2 = 0. Thus the generalized eigenvector can be chosen arbitrarily, except that it must be independent of (1) . c. Let = (0, 1) T . Then determine from the second of equations (37) and observe that = (1, 1) T = (1) . This choice of reproduces the solution found in Example 2. d. Let = (1, 0) T and determine the corresponding eigenvector . e. Let = (k1, k2) T , where k1 and k2 are arbitrary numbers. What condition on k1 and k2 ensures that and (1) are linearly independent? Then determine . How is it related to the eigenvector (1) ? 1